the derivative: $$\\frac{dy}{dx} = 8x - \\frac{1}{x^2}$$"},{"id":"block-2","content":"Next, set the derivative to zero and solve for $x$. Multiply by $x^2$ to clear the fraction and simplify the resulting cubic equation: $$8x - \\frac{1}{x^2} = 0 \\Rightarrow 8x^3 - 1 = 0 \\Rightarrow x^3 = \\frac{1}{8} \\Rightarrow x = \\frac{1}{2}$$"},{"id":"block-3","content":"Finally, substitute $x = \\frac{1}{2}$ into the original equation, not the derivative, to find the $y$-coordinate: $$y = 4\\left(\\frac{1}{2}\\right)^2 + \\frac{1}{\\frac{1}{2}} = 1 + 2 = 3$$ Therefore, the stationary point is $\\left(\\frac{1}{2}, 3\\right)$."}]},{"id":"card-10","type":"fill_blank","order":10,"blanks":[{"id":"x","isMath":true,"correctAnswer":"\\frac{1}{2}"},{"id":"y","isMath":true,"correctAnswer":"3"}],"sentence":"For the stationary point of $y = 4x^2 + \\frac{1}{x}$, the x-coordinate is {{blank:x}} and the y-coordinate is {{blank:y}}."},{"id":"card-11","type":"content","order":11,"title":"Where a curve has negative gradient","blocks":[{"id":"block-1","content":"To find where a curve has negative gradient, differentiate and solve $\\frac{dy}{dx} < 0$. For the function $y = x^3 - 3x^2 - 9x + 6$, we first calculate the derivative:"},{"id":"block-2","content":"$$$\\frac{dy}{dx} = 3x^2 - 6x - 9$$"},{"id":"block-3","content":"Next, we solve the inequality $3x^2 - 6x - 9 < 0$. Factorising the quadratic reveals its roots, and since the $x^2$ coefficient is positive, the graph opens upwards:"},{"id":"block-4","content":"$$$3x^2 - 6x - 9 < 0 \\Rightarrow 3(x - 3)(x + 1) < 0$$"},{"id":"block-5","content":"The solution to the inequality is $-1 < x < 3$. The endpoints are excluded from the result because the gradient is exactly zero at those specific points."}]},{"id":"card-12","type":"mcq","order":12,"options":[{"id":"a","text":"$$x < -2$ or $x > 4$","isCorrect":false},{"id":"b","text":"$$-2 < x < 4$","isCorrect":true},{"id":"c","text":"$$x \\le -2$ or $x \\ge 4$","isCorrect":false},{"id":"d","text":"$$x = -2$ or $x = 4$","isCorrect":false}],"question":"Stretch: If $\\frac{dy}{dx} = (x-4)(x+2)$, for which values of $x$ is the gradient negative?","explanation":"The roots are $x = -2$ and $x = 4$, and the positive coefficient means the quadratic is below the axis between the roots, so the gradient is negative on $-2 < x < 4$. A common mistake is A, which would be true for a downward-opening quadratic, and the endpoints are excluded because there the gradient is zero.","correctOptionId":"b"},{"id":"card-13","type":"flashcard","cards":[{"id":"fc-1","back":"$$v = \\frac{ds}{dt}$","front":"Velocity from displacement"},{"id":"fc-2","back":"$$a = \\frac{dv}{dt}$","front":"Acceleration from velocity"},{"id":"fc-3","back":"$$a = \\frac{d^2s}{dt^2}$","front":"Acceleration from displacement"},{"id":"fc-4","back":"$$v = 0$","front":"Instantaneously at rest"}],"order":13},{"id":"card-14","type":"content","order":14,"title":"Differentiation in motion","blocks":[{"id":"block-1","content":"In motion questions, displacement is usually $s$, velocity is usually $v$, and acceleration is usually $a$. If $s$ is given in terms of time $t$, then $v = \\frac{ds}{dt}$ and $a = \\frac{dv}{dt}$."},{"id":"block-2","content":"For $s = t^3 - 4t^2 - 3t + 6$, differentiate once to get velocity. A particle is instantaneously at rest when $v = 0$. $$v = 3t^2 - 8t - 3$$"},{"id":"block-3","content":"So solve $$3t^2 - 8t - 3 = 0 \\Rightarrow (3t + 1)(t - 3) = 0$$ giving $t = -\\frac{1}{3}$ or $t = 3$. If $t \\ge 0$, only $t = 3$ is valid, and differentiating again gives $$a = 6t - 8$$"}]},{"id":"card-15","type":"fill_blank","order":15,"blanks":[{"id":"t","isMath":true,"correctAnswer":"3"}],"sentence":"If $s = t^3 - 4t^2 - 3t + 6$ for $t \\ge 0$, the particle is instantaneously at rest at $t =$ {{blank:t}}."},{"id":"card-16","type":"multi-question","order":16,"questions":[{"id":"mq-1","options":[{"id":"a","text":"$$5x^4$","isCorrect":true},{"id":"b","text":"$$x^4$","isCorrect":false},{"id":"c","text":"$$5x^6$","isCorrect":false}],"question":"Differentiate $x^5$.","explanation":"Using the power

Differentiation Lesson